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Finding the Shape of a Funnel Curve


Date: 04/08/2001 at 02:43:36
From: James
Subject: Applied calculus

There is a circular coin donating receptacle I've noticed which has a 
small chute in which the coin is placed. When the coin leaves the 
chute it rotates around the inner surface of the receptacle, which 
is so shaped that the coin remains perpendicular to its surface as it 
rotates and gradually descends to the small aperture at the bottom of 
the inner surface. If you were to slice the receptacle vertically, the 
inner surface would look a bit like an inverted hyperboloid of 
revolution. 

Can anyone tell me how to determine the equation of the curve of 
revolution?


Date: 04/10/2001 at 13:36:29
From: Doctor Douglas
Subject: Re: Applied calculus

Hi James, and thanks for writing.

This is a very interesting question. I imagine that the receptacle you 
are referring to is shaped something like a smooth curved funnel.  
When you put a coin in the slot, it rolls around and around and 
gradually spirals inward until it drops out at the bottom.

If we slice the funnel vertically, the section (the cut edge) has a 
certain shape. Our task is to determine the shape, or equation, of 
this curve.

If the curve is a hyperbola (so that the funnel surface is a 
hyperboloid of revolution), then the equation for the curve is

  r^2   z^2
  --- - --- = 1    a generic hyperbola
  a^2   b^2

where I've written it in cylindrical coordinates (z is the vertical 
coordinate and r is the distance from the vertical central axis).  To 
specify the numbers a and b, we need to measure at least two points on
the curve. For example, we could measure the radius right where the
curve becomes vertical. This happens when z = 0, and thus at this 
point, a = r. We could also measure the asymptotic radial slope, which 
is the ratio z/r as r becomes very large at the outer edge. This slope 
will tend to the ratio b/a. It's also possible (although a bit 
messier) to compute the values for b and a by measuring points at 
other locations on the curve.

Unfortunately, we don't actually know whether or not the funnel is in 
fact a hyperboloid of revolution. It's possible that the funnel is a 
revolution of the curve z = C-1/r. This curve is used to model the 
gravitational potential well of a point mass, such as the sun, so that 
the funnel in this case will model the planets (coins, or marbles) 
rolling around the surface, going faster and faster at smaller radii.  
I've seen this type of apparatus used in science museums to illustrate 
this gravitational "potential well."  

Let's assume for the time being that these two shapes are the only 
choices. This brings us to another clever method for "measuring" the 
shape of the funnel: we time the orbits of the coin as it goes around.  
If the approximate period T(r), as a function of the radius r, goes as 

  T = k * sqrt(r^3)    Kepler law for "gravitational funnel"

then the curve is well approximated by z = C-1/r.  This is the 
statement of Kepler's third law, appropriate for objects that orbit in 
a gravitational potential of the form 1/r. It is only an approximation 
for the 1/r-shaped funnel, because the orbits of the coins are 
actually spirals.

What would the orbital period be for a hyperbola? In this case, the
gravitational potential is given by U = k*z = k b(r^2/a^2 - 1), for 
r > a. Using some simple physics (i.e., the conservation of energy), 
we can obtain the kinetic energy K of the object as a function of r

  K = K0 - U = K0 + kb(1 - r^2/a^2)

We choose the constant K0 such that the kinetic energy is zero at 
r = R, and then K = (k b/a^2)(R^2 - r^2). We then set K equal to 
m*v^2/2, and solve for the dependence of the orbital period T_hyp on 
r:

  T_hyp = 2*pi*r / v
        = 2*pi*r*sqrt(m a^2 / 2 k b)*sqrt(R^2 - r^2)
        = C r / sqrt(R^2 - r^2)            a < r < R, hyperbolic 
                                                       funnel

This is a different form than the Kepler law for the 1/r gravitational
funnel. It is much steeper out at large radii near r = R. Note that 
the hyperbola goes to a constant _slope_ (i.e., b/a) for large r, 
while the 1/r funnel goes to a constant for large r.  Hence the period 
of the coin should change much faster with radius at the outer edge in 
the case of the hyperbola, because as its radius decreases, it gains a 
lot of kinetic energy (from the slope) and its speed must increase 
dramatically, at least relative to the 1/r funnel.

So by measuring the dependence of the period T on the radius r, you 
can discriminate between the two curves above. It's rather neat that 
you don't need to make any measurements about the height z(r) to do 
this! Of course you need to measure r somehow, and you need to make 
the reasonable assumption that each "period" of an orbit is 
approximately a circle. And it's of course possible that the funnel is 
the revolution of a curve that is neither hyperbolic nor -1/r. By 
making many careful measurements of T(r) and using the assumptions 
above, it's still possible to extract the shape in this case, but at 
this point it might be prudent just to measure z(r) directly.

I hope this helps answer your question.  

- Doctor Douglas, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
College Calculus
College Physics
High School Calculus
High School Physics/Chemistry

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